October 2018 1 23 Report

A) -1 (x + 1) \leq 3 (x - 1) + 1

B) 4 - ułamek x - 2 (na górze) i pod kreską ułamkową dwa \leq - 3 (x + 5)

C) 2x - x - 1 na górze ułamka i na dole dwa \geq 3x + 2

D) - 4 (x - 3) + 6 (x - 1) = 10x + 2


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